What is the general solution for (cos(x))/(sin(2x))= 5/7 ?

The general solution for (cos(x))/(sin(2x))= 5/7 is x=0.77539…+2pin,x=pi-0.77539…+2pin

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(cos(x))/(sin(2x))= 5/7

\frac{cos ( x )}{sin ( 2 x )} = \frac{5}{7}

x = 0.77539 \dots + 2 πn, x = π - 0.77539 \dots + 2 πn

Solution steps

\frac{cos ( x )}{sin ( 2 x )} = \frac{5}{7}

Subtract

\frac{5}{7}

from both sides

\frac{cos ( x )}{sin ( 2 x )} - \frac{5}{7} = 0

Simplify

\frac{cos ( x )}{sin ( 2 x )} - \frac{5}{7} : \frac{7 cos ( x ) - 5 sin ( 2 x )}{7 sin ( 2 x )}

\frac{7 cos ( x ) - 5 sin ( 2 x )}{7 sin ( 2 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

7 cos (x) - 5 sin (2 x) = 0

Rewrite using trig identities

7 cos (x) - 10 cos (x) sin (x) = 0

Factor

7 cos (x) - 10 cos (x) sin (x) : - cos (x) (10 sin (x) - 7)

- cos (x) (10 sin (x) - 7) = 0

Solving each part separately

cos (x) = 0 or 10 sin (x) - 7 = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

10 sin (x) - 7 = 0 : x = arcsin (\frac{7}{10}) + 2 πn, x = π - arcsin (\frac{7}{10}) + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{7}{10}) + 2 πn, x = π - arcsin (\frac{7}{10}) + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

x = arcsin (\frac{7}{10}) + 2 πn, x = π - arcsin (\frac{7}{10}) + 2 πn

Show solutions in decimal form

x = 0.77539 \dots + 2 πn, x = π - 0.77539 \dots + 2 πn

tan (x) = \frac{6 \cdot 0.809}{15}

tan (θ) = \frac{8}{20}

cos^{2} (x) = \frac{( 5 k - 8 )}{5}

3 arcsin (x) = \frac{π}{2}

3 cos^{2} (x) - 1 = sin^{2} (x) + 2 sin (x)

What is the general solution for (cos(x))/(sin(2x))= 5/7 ?
The general solution for (cos(x))/(sin(2x))= 5/7 is x=0.77539…+2pin,x=pi-0.77539…+2pin